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Answer 1 · 2016-11-22T00:01:01

Start with $r = root(3)(GMT^2/(4pi^2))$

$r = root(3)(GMT^2/(4pi^2))$

Where r is the radius of the orbit, measured from the center of the Earth.

The distance, d, above the Earth is $r - R_(earth)$

$d = root(3)(GMT^2/(4pi^2)) - R_(earth)$

where $R_(earth)$ is the radius of the Earth.

Six hours converted to seconds:
$T = 21600 s$

Gravitational Constant
$G = 6.67 xx 10^-11 m^3kg^-1s^-2$

Earth's Mass
$M = 5.972xx10^24kg$

Earth's Radius
$R_(earth) = 6371000 m$

$d = root(3)((6.67 xx 10^-11 m^3kg^-1s^-2)(5.972xx10^24kg)(21600 s)^2/(4pi^2)) - 6371000 m$

$d = 10388631 m$